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For example, 3 × 5 is an integer factorization of 15, and (x – 2)(x + 2) is a polynomial factorization of x 2 – 4. Factorization is not usually considered meaningful within number systems possessing division , such as the real or complex numbers , since any x {\displaystyle x} can be trivially written as ( x y ) × ( 1 / y ) {\displaystyle ...
If one of these values is 0, we have a linear factor. If the values are nonzero, we can list the possible factorizations for each. Now, 2 can only factor as 1×2, 2×1, (−1)×(−2), or (−2)×(−1). Therefore, if a second degree integer polynomial factor exists, it must take one of the values p(0) = 1, 2, −1, or −2. and likewise for p(1).
The polynomial P = x 4 + 1 is irreducible over Q but not over any finite field. On any field extension of F 2, P = (x + 1) 4. On every other finite field, at least one of −1, 2 and −2 is a square, because the product of two non-squares is a square and so we have; If =, then = (+) ().
Squares are always congruent to 0, 1, 4, 5, 9, 16 modulo 20. The values repeat with each increase of a by 10. In this example, N is 17 mod 20, so subtracting 17 mod 20 (or adding 3), produces 3, 4, 7, 8, 12, and 19 modulo 20 for these values. It is apparent that only the 4 from this list can be a square.
First we compute 2P. We have s(P) = s(1,1) = 4, so the coordinates of 2P = (x ′, y ′) are x ′ = s 2 – 2x = 14 and y ′ = s(x – x ′) – y = 4(1 – 14) – 1 = –53, all numbers understood (mod n). Just to check that this 2P is indeed on the curve: (–53) 2 = 2809 = 14 3 + 5·14 – 5. Then we compute 3(2P). We have s(2P) = s(14 ...
The entry 4+2i = −i(1+i) 2 (2+i), for example, could also be written as 4+2i= (1+i) 2 (1−2i). The entries in the table resolve this ambiguity by the following convention: the factors are primes in the right complex half plane with absolute value of the real part larger than or equal to the absolute value of the imaginary part.
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For example, 15 is a composite number because 15 = 3 · 5, but 7 is a prime number because it cannot be decomposed in this way. If one of the factors is composite, it can in turn be written as a product of smaller factors, for example 60 = 3 · 20 = 3 · (5 · 4).